Run function in com_content with ajax - doesn't call it corretly

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1
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I've been following this tutorial to make an AJAX login form.
http://www.mysite4u.net/joomla-articles/jquery-ajax-joomla-login.html

But I can't get it to work. The ajax call dosen't work. It gives me this error Uncaught TypeError: Cannot read property 'loggedIn' of null

I tried reading this documentation, but I guess that it is not the AJAX component that is used in this, and it dosen't mention anything about calling components: https://docs.joomla.org/Using_Joomla_Ajax_Interface

This is my AJAX call:

form = jQuery('#userForm'); 

jQuery.ajax(
 url: 'index.php?option=com_content&format=ajax&view=article&task=loginUser&tmpl=component',
 type: 'post',
 dataType: 'html',
 data: 
 form: jQuery(form).serializeArray()
 ,
 async: true,
 success: function(response)
 console.log(response);
 var result = jQuery.parseJSON(response);

 if ( result.loggedIn == 1 ) 
 //User logged in 
 
 else 
 //User not logged in 
 
 
);

And the function inside components/com_content/views/article/view.ajax.php
The "view.ajax.php" is a file I've created:

public function _logInUser( $form )
 $options = array();
 $credentials = array();

 $credentials['username'] = $form[0]['value'];
 $credentials['password'] = $form[1]['value'];

 $result = JFactory::getApplication()->login($credentials, $options); 
 $result = ($result) ? 1 : 0;

 //1 - logged in
 //0 - not logged in 
 echo json_encode( array('loggedIn' => $result) ); 
 jexit();

edited 30 mins ago

asked 1 hour ago

JonasB

1608

add a commentÂ |Â

up vote
1
down vote

favorite

I've been following this tutorial to make an AJAX login form.
http://www.mysite4u.net/joomla-articles/jquery-ajax-joomla-login.html

But I can't get it to work. The ajax call dosen't work. It gives me this error Uncaught TypeError: Cannot read property 'loggedIn' of null

This is my AJAX call:

form = jQuery('#userForm'); 

jQuery.ajax(
 url: 'index.php?option=com_content&format=ajax&view=article&task=loginUser&tmpl=component',
 type: 'post',
 dataType: 'html',
 data: 
 form: jQuery(form).serializeArray()
 ,
 async: true,
 success: function(response)
 console.log(response);
 var result = jQuery.parseJSON(response);

 if ( result.loggedIn == 1 ) 
 //User logged in 
 
 else 
 //User not logged in 
 
 
);

And the function inside components/com_content/views/article/view.ajax.php
The "view.ajax.php" is a file I've created:

public function _logInUser( $form )
 $options = array();
 $credentials = array();

 $credentials['username'] = $form[0]['value'];
 $credentials['password'] = $form[1]['value'];

 $result = JFactory::getApplication()->login($credentials, $options); 
 $result = ($result) ? 1 : 0;

 //1 - logged in
 //0 - not logged in 
 echo json_encode( array('loggedIn' => $result) ); 
 jexit();

edited 30 mins ago

asked 1 hour ago

JonasB

1608

add a commentÂ |Â

up vote
1
down vote

favorite

I've been following this tutorial to make an AJAX login form.
http://www.mysite4u.net/joomla-articles/jquery-ajax-joomla-login.html

But I can't get it to work. The ajax call dosen't work. It gives me this error Uncaught TypeError: Cannot read property 'loggedIn' of null

This is my AJAX call:

form = jQuery('#userForm'); 

jQuery.ajax(
 url: 'index.php?option=com_content&format=ajax&view=article&task=loginUser&tmpl=component',
 type: 'post',
 dataType: 'html',
 data: 
 form: jQuery(form).serializeArray()
 ,
 async: true,
 success: function(response)
 console.log(response);
 var result = jQuery.parseJSON(response);

 if ( result.loggedIn == 1 ) 
 //User logged in 
 
 else 
 //User not logged in 
 
 
);

And the function inside components/com_content/views/article/view.ajax.php
The "view.ajax.php" is a file I've created:

public function _logInUser( $form )
 $options = array();
 $credentials = array();

 $credentials['username'] = $form[0]['value'];
 $credentials['password'] = $form[1]['value'];

 $result = JFactory::getApplication()->login($credentials, $options); 
 $result = ($result) ? 1 : 0;

 //1 - logged in
 //0 - not logged in 
 echo json_encode( array('loggedIn' => $result) ); 
 jexit();

edited 30 mins ago

asked 1 hour ago

JonasB

1608

I've been following this tutorial to make an AJAX login form.
http://www.mysite4u.net/joomla-articles/jquery-ajax-joomla-login.html

But I can't get it to work. The ajax call dosen't work. It gives me this error Uncaught TypeError: Cannot read property 'loggedIn' of null

This is my AJAX call:

form = jQuery('#userForm'); 

jQuery.ajax(
 url: 'index.php?option=com_content&format=ajax&view=article&task=loginUser&tmpl=component',
 type: 'post',
 dataType: 'html',
 data: 
 form: jQuery(form).serializeArray()
 ,
 async: true,
 success: function(response)
 console.log(response);
 var result = jQuery.parseJSON(response);

 if ( result.loggedIn == 1 ) 
 //User logged in 
 
 else 
 //User not logged in 
 
 
);

And the function inside components/com_content/views/article/view.ajax.php
The "view.ajax.php" is a file I've created:

public function _logInUser( $form )
 $options = array();
 $credentials = array();

 $credentials['username'] = $form[0]['value'];
 $credentials['password'] = $form[1]['value'];

 $result = JFactory::getApplication()->login($credentials, $options); 
 $result = ($result) ? 1 : 0;

 //1 - logged in
 //0 - not logged in 
 echo json_encode( array('loggedIn' => $result) ); 
 jexit();

joomla-3.x ajax

edited 30 mins ago

asked 1 hour ago

JonasB

1608

edited 30 mins ago

asked 1 hour ago

JonasB

1608

edited 30 mins ago

asked 1 hour ago

JonasB

1608

asked 1 hour ago

JonasB

1608

asked 1 hour ago

JonasB

1608

add a commentÂ |Â

2 Answers
2

active

oldest

votes

up vote
1
down vote

Apperently the tutorial was missing an important part. It needed to have the "loginUser" task addded to the switch in view.ajax.php like:

function display($tpl = null)
 
 $task = JRequest::getString('task'); 
 switch( $task ) 
 case "registerNewUser": 
 $this->_registerNewUser( JFactory::getApplication()->input->post->get('form', NULL, 'ARRAY') ); 
 break;
 case "loginUser": 
 $this->_logInUser( JFactory::getApplication()->input->post->get('form', NULL, 'ARRAY') ); 
 break;
 
 jexit();

answered 20 mins ago

JonasB

1608

add a commentÂ |Â

up vote
1
down vote

Maybe I misunderstand you in your question but in the directory of com_content/views/article you cannot/should not create a view.html.php file since it originally has a good view.html.php file.

In the tutorial which you follow the author says that you should create an 'view.ajax.php' file in com_content/views/article folder (not a view.html.php file). Maybe that is causing your problem since you created a wrong file for this ajax call.

And if that is just a mistake in your question, then you also might want to study this earlier answer about ajax calls within Joomla here (it is better not to repeat that here):
What is the proper way to make an AJAX call in component?

edited 20 mins ago

answered 48 mins ago

Zollie

565

Woops, my fault... It is a view.ajax.php I've created. Thanks for your suggestion!
â€“Â JonasB
29 mins ago

I added a link in my answer about the proper ajax calls in Joomla, if you did not study that before.
â€“Â Zollie
17 mins ago

add a commentÂ |Â

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2 Answers
2

active

oldest

votes

2 Answers
2

active

oldest

votes

up vote
1
down vote

Apperently the tutorial was missing an important part. It needed to have the "loginUser" task addded to the switch in view.ajax.php like:

function display($tpl = null)
 
 $task = JRequest::getString('task'); 
 switch( $task ) 
 case "registerNewUser": 
 $this->_registerNewUser( JFactory::getApplication()->input->post->get('form', NULL, 'ARRAY') ); 
 break;
 case "loginUser": 
 $this->_logInUser( JFactory::getApplication()->input->post->get('form', NULL, 'ARRAY') ); 
 break;
 
 jexit();

answered 20 mins ago

JonasB

1608

add a commentÂ |Â

up vote
1
down vote

Apperently the tutorial was missing an important part. It needed to have the "loginUser" task addded to the switch in view.ajax.php like:

function display($tpl = null)
 
 $task = JRequest::getString('task'); 
 switch( $task ) 
 case "registerNewUser": 
 $this->_registerNewUser( JFactory::getApplication()->input->post->get('form', NULL, 'ARRAY') ); 
 break;
 case "loginUser": 
 $this->_logInUser( JFactory::getApplication()->input->post->get('form', NULL, 'ARRAY') ); 
 break;
 
 jexit();

answered 20 mins ago

JonasB

1608

add a commentÂ |Â

up vote
1
down vote

Apperently the tutorial was missing an important part. It needed to have the "loginUser" task addded to the switch in view.ajax.php like:

function display($tpl = null)
 
 $task = JRequest::getString('task'); 
 switch( $task ) 
 case "registerNewUser": 
 $this->_registerNewUser( JFactory::getApplication()->input->post->get('form', NULL, 'ARRAY') ); 
 break;
 case "loginUser": 
 $this->_logInUser( JFactory::getApplication()->input->post->get('form', NULL, 'ARRAY') ); 
 break;
 
 jexit();

answered 20 mins ago

JonasB

1608

Apperently the tutorial was missing an important part. It needed to have the "loginUser" task addded to the switch in view.ajax.php like:

function display($tpl = null)
 
 $task = JRequest::getString('task'); 
 switch( $task ) 
 case "registerNewUser": 
 $this->_registerNewUser( JFactory::getApplication()->input->post->get('form', NULL, 'ARRAY') ); 
 break;
 case "loginUser": 
 $this->_logInUser( JFactory::getApplication()->input->post->get('form', NULL, 'ARRAY') ); 
 break;
 
 jexit();

answered 20 mins ago

JonasB

1608

answered 20 mins ago

JonasB

1608

answered 20 mins ago

JonasB

1608

answered 20 mins ago

JonasB

1608

add a commentÂ |Â

up vote
1
down vote

Maybe I misunderstand you in your question but in the directory of com_content/views/article you cannot/should not create a view.html.php file since it originally has a good view.html.php file.

edited 20 mins ago

answered 48 mins ago

Zollie

565

Woops, my fault... It is a view.ajax.php I've created. Thanks for your suggestion!
â€“Â JonasB
29 mins ago

I added a link in my answer about the proper ajax calls in Joomla, if you did not study that before.
â€“Â Zollie
17 mins ago

add a commentÂ |Â

up vote
1
down vote

Maybe I misunderstand you in your question but in the directory of com_content/views/article you cannot/should not create a view.html.php file since it originally has a good view.html.php file.

edited 20 mins ago

answered 48 mins ago

Zollie

565

Woops, my fault... It is a view.ajax.php I've created. Thanks for your suggestion!
â€“Â JonasB
29 mins ago

I added a link in my answer about the proper ajax calls in Joomla, if you did not study that before.
â€“Â Zollie
17 mins ago

add a commentÂ |Â

up vote
1
down vote

Maybe I misunderstand you in your question but in the directory of com_content/views/article you cannot/should not create a view.html.php file since it originally has a good view.html.php file.

edited 20 mins ago

answered 48 mins ago

Zollie

565

Maybe I misunderstand you in your question but in the directory of com_content/views/article you cannot/should not create a view.html.php file since it originally has a good view.html.php file.

edited 20 mins ago

answered 48 mins ago

Zollie

565

edited 20 mins ago

answered 48 mins ago

Zollie

565

answered 48 mins ago

Zollie

565

answered 48 mins ago

Zollie

565

Woops, my fault... It is a view.ajax.php I've created. Thanks for your suggestion!
â€“Â JonasB
29 mins ago

I added a link in my answer about the proper ajax calls in Joomla, if you did not study that before.
â€“Â Zollie
17 mins ago

add a commentÂ |Â

Woops, my fault... It is a view.ajax.php I've created. Thanks for your suggestion!
â€“Â JonasB
29 mins ago

I added a link in my answer about the proper ajax calls in Joomla, if you did not study that before.
â€“Â Zollie
17 mins ago

Woops, my fault... It is a view.ajax.php I've created. Thanks for your suggestion!
â€“Â JonasB
29 mins ago

I added a link in my answer about the proper ajax calls in Joomla, if you did not study that before.
â€“Â Zollie
17 mins ago

add a commentÂ |Â

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